= ( 1  ) ( 2  )  
1  1  0   
0  1  1  
0  0  5    

= ( 1  ) ( 2  ) ( 5  )  
1  1  0    = 0 
0  1  1  
0  0  1   

Take ( 5  ) from the 3rd row as a coefficient 

 

The determinant of the upper triangular matrix 
is = 1·1·1 = 1 
 

=> _{1} = 1, _{2} = 2, _{3} = 5 (eigenvalues)
=> We have
D =  
1  0  0  
0  2  0 
0  0  5  

Eigenvectors:
Using the equation (A  I ) = 0,
we find the eigenvectors of A:
_{1} = 1: _{1} = t ( 0, 1, 2 ), t C, t 0
_{2} = 2: _{2} = t ( 1, 1, 2 ), t C, t 0
_{3} = 6: _{3} = t ( 1, 1, 1 ), t C, r 0
Choose _{i}'s as follows:
_{1} = ( 0, 1, 2 ), _{2} = ( 1, 1, 2 ), _{3} = ( 1, 1, 1 )
and form the matrix T:
T =  
0  1  1  
1  1  1 
2  2  1  

We have
T^{ 1} =  
3  3  0  
3  2  1 
0  2  1  

so we now calculate
Te^{D}T^{ 1} =  
0  1  1  
1  1  1 
2  2  1  


e^{1}  0  0  
0  e^{2}  0 
0  0  e^{5}  


3  3  0  
3  2  1 
0  2  1  

=  
3e^{2}  2e^{2} + 2e^{5}  e^{2} + e^{5}  
3e  3e^{2}  3e + 2e^{2} + 2e^{5}  e^{2} + e^{5} 
6e  6e^{2}  6e  4e^{2}  2e^{5}  2e^{2} + e^{5}  

 
7,4  94,0  47,0  
4,7  96,7  47,0 
9,3  103,4  54,4  

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